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Thomas Calculus 13th [Solutions]

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796 Chapter 10 Infinite Sequences and Series<br />

3. diverges by the nth-Term Test since<br />

exist<br />

n n<br />

lim lim ( 1) tanh lim ( 1) 1<br />

n<br />

a e<br />

n<br />

n lim ( 1) does not<br />

2n<br />

n n b 1 e n<br />

2n<br />

n<br />

ln n!<br />

4. converges by the Direct Comparison Test: n! n ln n! n ln ( n) ln ( n)<br />

n log n n!<br />

n<br />

log n n ! 1 , which is the nth-term of a convergent p-series<br />

3 2<br />

n n<br />

5. converges by the Direct Comparison Test: a 12 1 2 12 2 3 1 2 12<br />

1 a<br />

2 2 3 4 a<br />

2 3 4 5 3 4<br />

2<br />

(1)(3)(2) (2)(4)(3) (3)(5)(4)<br />

a 4 3 4 2 3 1 2 12 , 1<br />

12<br />

5 6 4 5 3 4 2<br />

(4)(6)(5) ( n 1)( n 3)( n<br />

n 1<br />

represents the given series and<br />

2<br />

2)<br />

12 12 , which is the nth-term of a convergent p-series<br />

2 4<br />

( n 1)( n 3)( n 2) n<br />

an<br />

1<br />

6. converges by the Ratio Test: lim lim n 0 1<br />

n<br />

an<br />

n<br />

( n 1)( n 1)<br />

7. diverges by the nth-Term Test since if an<br />

L as n , then<br />

1 2 1 5<br />

L L L 1 0 L 0<br />

1 L<br />

2<br />

8. Split the given series into 1 and 2n<br />

; the first subseries is a convergent geometric series and the<br />

2n<br />

1<br />

2n 3<br />

3<br />

n 1<br />

n 1<br />

second converges by the Root Test: lim 2 lim n n<br />

n n 2 n 11 1<br />

2<br />

3<br />

9 9 9<br />

1<br />

n<br />

n<br />

n<br />

3 3 (4)<br />

9. f ( x) cos x with a f 0.5, f , f 0.5, f , f 0.5;<br />

3 3 3 2 3 3 2 3<br />

3<br />

2<br />

3<br />

3<br />

cos x 1 x 1 x x<br />

2 2 3 4 3 12 3<br />

10. f ( x) sin x with<br />

(4) (5)<br />

a 2 f (2 ) 0, f (2 ) 1, f (2 ) 0, f (2 ) 1, f (2 ) 0, f (2 ) 1,<br />

(6) (7)<br />

f (2 ) 0, f (2 ) 1; sin x ( x 2 )<br />

3 5 7<br />

( x 2 ) ( x 2 ) ( x 2 )<br />

3! 5! 7!<br />

11.<br />

2 3<br />

x<br />

e 1 x x x with a 0<br />

2! 3!<br />

12. f ( x) ln x with<br />

ln x ( x 1)<br />

13. f ( x) cos x with<br />

(4)<br />

a 1 f (1) 0, f (1) 1, f (1) 1, f (1) 2, f (1) 6;<br />

2 3 4<br />

( x 1) ( x 1) ( x 1)<br />

2 3 4<br />

(5) (6)<br />

f (22 ) 0, f (22 ) 1;<br />

(4)<br />

a 22 f (22 ) 1, f (22 ) 0, f (22 ) 1, f (22 ) 0, f (22 ) 1,<br />

1 2 1 4 1<br />

6<br />

cos x 1 ( x 22 ) ( x 22 ) ( x 22 )<br />

2 4! 6!<br />

Copyright<br />

2014 Pearson Education, Inc.

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